擅长:python、mysql、java
<p>您可以使用<code>itertools.groupby</code>:</p>
<pre><code>from itertools import groupby
lst = ['S1', 'S2', 'S6', 'S1', 'S2', 'S3', 'S4', 'S5', 'S1', 'S2', 'S5', 'S1', 'S2', 'S4', 'S5', 'S1', 'S2', 'S4', 'S5', 'S1', 'S2', 'S3', 'S6']
splitby = 'S1'
res = [[splitby] + list(g) for k, g in groupby(lst, key=lambda x: x != splitby) if k]
# [['S1', 'S2', 'S6'], ['S1', 'S2', 'S3', 'S4', 'S5'], ['S1', 'S2', 'S5'], ['S1', 'S2', 'S4', 'S5'], ['S1', 'S2', 'S4', 'S5'], ['S1', 'S2', 'S3', 'S6']]
</code></pre>